answer:1. **Identify the roots**: Let the roots of the polynomial ( P(x) = x^k + a_1 x^{k-1} + a_2 x^{k-2} + ldots + a_k ) be (alpha_1, alpha_2, ldots, alpha_k ). 2. **Sum of the roots**: By Vieta's formulas, the sum of the roots taken one at a time (with sign consideration) satisfies: [ alpha_1 + alpha_2 + ldots + alpha_k = -a_1 ] 3. **Sum of the product of roots taken two at a time**: Similarly, the sum of the products of the roots taken two at a time satisfies: [ sum_{1 leq i < j leq k} alpha_i alpha_j = a_2 ] 4. **Expression involving root differences**: Consider the expression (sum_{i neq j} (alpha_i - alpha_j)^2). Since the roots are distinct and real, we want to show this sum is greater than 0: [ sum_{i neq j} (alpha_i - alpha_j)^2 > 0 ] Expanding this, we get: [ sum_{i neq j} (alpha_i^2 - 2alpha_i alpha_j + alpha_j^2) ] 5. **Simplify the sum**: Each term (alpha_i^2) will appear (k-1) times. Thus, the above expression becomes: [ (k-1) sum_{i=1}^k alpha_i^2 - 2 sum_{1 leq i < j leq k} alpha_i alpha_j ] 6. **Relation to (a_2) and (a_1)**: Using the fact that: [ left( sum_{i=1}^k alpha_i right)^2 = sum_{i=1}^k alpha_i^2 + 2 sum_{1 leq i < j leq k} alpha_i alpha_j ] Substitute: [ (-a_1)^2 = sum_{i=1}^k alpha_i^2 + 2a_2 ] Therefore: [ a_1^2 = sum_{i=1}^k alpha_i^2 + 2a_2 ] 7. **Substitute back into the initial sum**: We finally get: [ sum_{i neq j} (alpha_i - alpha_j)^2 = (k-1) sum_{i=1}^k alpha_i^2 - 2 sum_{1 leq i < j leq k} alpha_i alpha_j ] Substituting the above relationship: [ (k-1) (sum_{i=1}^k alpha_i^2) - 2a_2 = (k-1) left( a_1^2 - 2a_2 right) - 2a_2 ] 8. **Simplifying inequality**: Therefore: [ (k-1) a_1^2 - (2k - 2)a_2 > 0 ] This simplifies to: [ (k-1)a_1^2 > 2ka_2 ] 9. **Conclusion**: Hence, we have successfully demonstrated that: [ a_1^2 > frac{2ka_2}{k-1} ] [ boxed{a_1^2 > frac{2ka_2}{k-1}} ]

answer:Let's break down the solution into detailed steps for each option: **Option A:** Given f(x) = 1 - frac{1}{x} - ln x, we find its derivative to analyze the function's behavior: 1. Calculate the derivative: f'(x) = frac{1}{x^2} - frac{1}{x}. 2. Analyze f'(x): - When 0 < x < 1, f'(x) > 0 because the 1/x^2 term dominates, making the derivative positive. - When x > 1, f'(x) < 0 since -frac{1}{x} becomes the dominant term, making the derivative negative. 3. At x = 1, f(x) = 1 - 1 - ln 1 = 0, which is its maximum value. 4. Therefore, f(x) leq 0 for all x > 0, which implies 1 - frac{1}{x} leq ln x. Hence, option A is boxed{text{correct}}. **Option B:** 1. Let x = frac{pi}{4}, then sin 2x = sin frac{pi}{2} = 1. 2. Since sin frac{pi}{2} = 1 > frac{pi}{4}, the inequality does not hold for all x > 0. 3. Thus, option B is boxed{text{incorrect}}. **Option C:** 1. Simplify the given expression: frac{1 + tan frac{pi}{12}}{1 - tan frac{pi}{12}} = tanleft(frac{pi}{4} + frac{pi}{12}right) = tanfrac{pi}{3}. 2. Since tanfrac{pi}{3} > frac{pi}{3} (as the tangent function is increasing in the first quadrant and tanfrac{pi}{4} = 1 < frac{pi}{3}), the inequality holds. 3. Therefore, option C is boxed{text{correct}}. **Option D:** 1. Define h(x) = x - sin x and find its derivative: h'(x) = 1 - cos x geq 0 for x > 0. 2. Since h'(x) geq 0, h(x) is monotonically increasing, implying x > sin x for x > 0. 3. To show e^x > 2sin x, it suffices to prove e^x > 2x: - Let g(x) = e^x - 2x and find its derivative: g'(x) = e^x - 2. - Setting g'(x) = 0 gives x = ln 2, which is the point where g(x) changes its monotonicity. - g(x) is decreasing for 0 < x < ln 2 and increasing for x > ln 2. The minimum value of g(x) occurs at x = ln 2: g(ln 2) = 2(1 - ln 2) > 0. 4. Since g(x) geq g(ln 2) > 0 for x > 0, we have e^x > 2x and thus e^x > 2sin x. 5. Hence, option D is boxed{text{correct}}. Therefore, the correct options are boxed{ACD}.

answer:(1) Given b_1 + b_3 = 5 and b_1b_3 = 4, and since b_1 < b_3, we can solve for b_1 and b_3. Then we can find the common ratio q. Inserting these values into the formula for a geometric progression, we can solve for the general term. First, because {b_n} is a geometric progression, we can express b_3 as b_1q^2. Now we substitute b_3 with b_1q^2 in our equations: b_1 + b_1q^2 = 5, b_1^2q^2 = 4. From the second equation, we find b_1^2 = frac{4}{q^2}. Substituting this into the first equation gives us: sqrt{frac{4}{q^2}} + (sqrt{frac{4}{q^2}})q^2 = 5. Simplifying, we get: 2left(frac{1}{q} + qright) = 5, leading to: q + frac{1}{q} = frac{5}{2}. This is a quadratic equation in terms of q. Multiplying both sides by q gives us: q^2 - frac{5}{2}q + 1 = 0. Solving this quadratic equation for q, we get two possible values for q (since b_n is an increasing sequence, we only take the positive value): q = frac{frac{5}{2} pm sqrt{left(frac{5}{2}right)^2 - 4}}{2}. Taking positive value of q corresponding to an increasing sequence, we find: q = frac{5 + sqrt{17}}{4}. Now we solve for b_1. Using b_1b_3 = 4, we have b_1^2q^2 = 4, and hence: b_1 = sqrt{frac{4}{q^2}}. After inserting the value of q, we get: b_1 = frac{2}{q} = frac{2}{frac{5 + sqrt{17}}{4}} = frac{8}{5 + sqrt{17}}. By rationalizing the denominator, we finally find: b_1 = frac{8}{5 + sqrt{17}} cdot frac{5 - sqrt{17}}{5 - sqrt{17}} = frac{8(5 - sqrt{17})}{25 - 17} = 5 - sqrt{17}. The general formula for a geometric series is b_n = b_1q^{n-1}, so we get: b_n = (5 - sqrt{17})left(frac{5 + sqrt{17}}{4}right)^{n-1}. And our boxed answer: boxed{b_n = (5 - sqrt{17})left(frac{5 + sqrt{17}}{4}right)^{n-1}}. (2) For the sequence {a_n} = log_2 b_n + 3 to be an arithmetic progression, it must have a constant difference between consecutive terms, which means we need to show that a_{n+1} - a_n = d (where d is a constant). Substituting b_n = (5 - sqrt{17})left(frac{5 + sqrt{17}}{4}right)^{n-1} into a_n, we have: a_n = log_2left[(5 - sqrt{17})left(frac{5 + sqrt{17}}{4}right)^{n-1}right] + 3. We separate the logarithm into two parts: a_n = log_2(5 - sqrt{17}) + (n-1)log_2left(frac{5 + sqrt{17}}{4}right) + 3. Notice that the first term log_2(5 - sqrt{17}) is a constant, and the second term is (n-1) times a constant, therefore the difference: a_{n+1} - a_n = left[log_2(5 - sqrt{17}) + nlog_2left(frac{5 + sqrt{17}}{4}right) + 3right] - left[log_2(5 - sqrt{17}) + (n-1)log_2left(frac{5 + sqrt{17}}{4}right) + 3right], simplifies to: a_{n+1} - a_n = log_2left(frac{5 + sqrt{17}}{4}right). Since log_2left(frac{5 + sqrt{17}}{4}right) is a constant, it proves that {a_n} is an arithmetic sequence with common difference d = log_2left(frac{5 + sqrt{17}}{4}right), and we can box this as our final statement regarding the arithmetic sequence: boxed{a_n text{ is an arithmetic sequence with } d = log_2left(frac{5 + sqrt{17}}{4}right)}.

answer:The time taken for two trains to cross each other when moving in opposite directions is the time taken for them to cover the sum of their lengths at their relative speed. Let's denote the length of the slower train as L1 and the length of the faster train as L2. According to the problem, L1 is 500 m. The relative speed of the two trains when moving in opposite directions is the sum of their individual speeds. So, the relative speed (V_rel) is: V_rel = 100 kmph + 120 kmph V_rel = 220 kmph To work with meters and seconds, we need to convert the relative speed from kmph to m/s. We use the conversion factor 1 km/h = 1000 m / 3600 s. V_rel (in m/s) = 220 * (1000 m / 3600 s) V_rel (in m/s) = 220 * (10/36) V_rel (in m/s) = 220 * (5/18) V_rel (in m/s) = 61.111... m/s The time taken (t) for the slower train to cross the faster train is given as 19.6347928529354 seconds. The total distance covered (D) when the slower train crosses the faster train is the sum of their lengths: D = L1 + L2 We can use the formula for time, distance, and speed: t = D / V_rel Plugging in the values we have: 19.6347928529354 s = (500 m + L2) / 61.111... m/s Now we solve for L2: 19.6347928529354 s * 61.111... m/s = 500 m + L2 1200 m (approximately) = 500 m + L2 L2 = 1200 m - 500 m L2 = 700 m So, the length of the faster train (L2) is boxed{700} meters.